Theorem sbalex 2278

Description: Equivalence of two ways to express proper substitution of a setvar for another setvar disjoint from it in a formula. This proof of their equivalence does not use df-sb 2097.

That both sides of the biconditional express proper substitution is proved by sb5 2311 and sb6 2119. The implication "to the left" is equs4v 2030 and does not require ax-10 2176 nor ax-12 2213. It also holds without disjoint variable condition if we allow more axioms (see equs4 2448). Theorem 6.2 of [Quine] p. 40. Theorem equs5 2492 replaces the disjoint variable condition with a distinctor antecedent. Theorem equs45f 2491 replaces the disjoint variable condition on 𝑥, 𝑡 with the nonfreeness hypothesis of 𝑡 in 𝜑. (Contributed by NM, 14-Apr-2008.) Revised to use equsexv 2304 in place of equsex 2450 in order to remove dependency on ax-13 2404. (Revised by BJ, 20-Dec-2020.) Revise to remove dependency on df-sb 2097. (Revised by BJ, 21-Sep-2024.) (Proof shortened by SN, 14-Aug-2025.)

Assertion

Ref	Expression
sbalex	⊢ (∃𝑥(𝑥 = 𝑡 ∧ 𝜑) ↔ ∀𝑥(𝑥 = 𝑡 → 𝜑))

Distinct variable group:   𝑥,𝑡

Allowed substitution hints:   𝜑(𝑥,𝑡)

Proof of Theorem sbalex

Step	Hyp	Ref	Expression
1		nfe1 2185	. . 3 ⊢ Ⅎ𝑥∃𝑥(𝑥 = 𝑡 ∧ 𝜑)
2		ax12ev2 2216	. . 3 ⊢ (∃𝑥(𝑥 = 𝑡 ∧ 𝜑) → (𝑥 = 𝑡 → 𝜑))
3	1, 2	alrimi 2249	. 2 ⊢ (∃𝑥(𝑥 = 𝑡 ∧ 𝜑) → ∀𝑥(𝑥 = 𝑡 → 𝜑))
4		equs4v 2030	. 2 ⊢ (∀𝑥(𝑥 = 𝑡 → 𝜑) → ∃𝑥(𝑥 = 𝑡 ∧ 𝜑))
5	3, 4	impbii 212	1 ⊢ (∃𝑥(𝑥 = 𝑡 ∧ 𝜑) ↔ ∀𝑥(𝑥 = 𝑡 → 𝜑))

Syntax hints:   → wi 4   ↔ wb 209   ∧ wa 400  ∀wal 1568  ∃wex 1809

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1825  ax-4 1839  ax-5 1940  ax-6 1997  ax-7 2038  ax-10 2176  ax-12 2213

This theorem depends on definitions:  df-bi 210  df-an 401  df-ex 1810  df-nf 1814

This theorem is referenced by:  sb5  2311  dfsb7  2314  alexeqg  3611  dfdif3OLD  4074  regsfromsetind  37028  pm13.196a  45104